Answer
$$y = \frac{1}{9}{\sec ^3}3x - \frac{1}{3}\sec 3x + C$$
Work Step by Step
$$\eqalign{
& y' = {\tan ^3}3x\sec 3x \cr
& {\text{Write }}y'{\text{ as }}\frac{{dy}}{{dx}} \cr
& \frac{{dy}}{{dx}} = {\tan ^3}3x\sec 3x \cr
& {\text{Separate the variables}} \cr
& dy = {\tan ^3}3x\sec 3xdx \cr
& {\text{Integrate both sides}} \cr
& \int {dy} = \int {{{\tan }^3}3x\sec 3xdx} \cr
& y = \int {{{\tan }^2}3x\tan 3x\sec 3xdx} \cr
& {\text{Use the identity }}{\sec ^2}\theta = 1 + {\tan ^2}\theta \cr
& y = \int {\left( {{{\sec }^2}3x - 1} \right)\tan 3x\sec 3xdx} \cr
& y = \frac{1}{3}\int {\left( {{{\sec }^2}3x} \right)\tan 3x\sec 3x\left( 3 \right)dx} - \frac{1}{3}\int {\tan 3x\sec 3x\left( 3 \right)dx} \cr
& {\text{Integrating}} \cr
& y = \frac{1}{3}\left( {\frac{{{{\sec }^3}3x}}{3}} \right) - \frac{1}{3}\sec 3x + C \cr
& y = \frac{1}{9}{\sec ^3}3x - \frac{1}{3}\sec 3x + C \cr} $$
